Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two.

Explanation

To find the magnitude of the gravitational force on any of the spheres due to the other two, we can follow these steps:

### Step 1: Understand the Configuration
We have three uniform spheres, each with mass \( M \) and radius \( a \), arranged such that each sphere touches the other two. The centers of the spheres form an equilateral triangle with each side equal to \( 2a \).

### Step 2: Determine the Gravitational Force Between Two Spheres
The gravitational force \( F \) between two point masses is given by Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
In our case, the mass \( m_1 \) and \( m_2 \) are both \( M \), and the distance \( r \) between the centers of any two spheres is \( 2a \). Therefore, the gravitational force between any two spheres (say sphere O and sphere A) is:
\[ F_{OA} = \frac{G M^2}{(2a)^2} = \frac{G M^2}{4a^2} \]

### Step 3: Calculate the Forces Acting on Sphere O
The gravitational force on sphere O due to sphere A is \( F_{OA} \) and due to sphere B is \( F_{OB} \). Both forces have the same magnitude:
\[ F_{OA} = F_{OB} = \frac{G M^2}{4a^2} \]

### Step 4: Determine the Angle Between Forces
Since the spheres form an equilateral triangle, the angle between the forces \( F_{OA} \) and \( F_{OB} \) is \( 60^\circ \).

### Step 5: Resolve the Forces into Components
To find the resultant force on sphere O, we resolve the forces into their components. The horizontal components of the forces will add up, while the vertical components will cancel out.

- The horizontal component of \( F_{OA} \):
\[ F_{OA,x} = F_{OA} \cos(30^\circ) = \frac{G M^2}{4a^2} \cdot \frac{\sqrt{3}}{2} \]
- The horizontal component of \( F_{OB} \):
\[ F_{OB,x} = F_{OB} \cos(30^\circ) = \frac{G M^2}{4a^2} \cdot \frac{\sqrt{3}}{2} \]

### Step 6: Calculate the Resultant Force
The net horizontal force \( F_{net} \) acting on sphere O is the sum of the horizontal components of \( F_{OA} \) and \( F_{OB} \):
\[ F_{net} = F_{OA,x} + F_{OB,x} = 2 \cdot \left( \frac{G M^2}{4a^2} \cdot \frac{\sqrt{3}}{2} \right) \]
\[ F_{net} = \frac{G M^2 \sqrt{3}}{4a^2} \]

### Final Result
Thus, the magnitude of the gravitational force on any of the spheres due to the other two is:
\[ F_{net} = \frac{\sqrt{3} G M^2}{4a^2} \]